Nakayama's Lemma

R. Virk

Department of Mathematics
University of Wisconsin
Madison, WI 53706

virk@math.wisc.edu

Lemma 0.0.1   Let $M$ be a finitely generated module over a commutative ring $A$ (with $1$). Suppose there is an ideal of $A$ such that $M=IM$. Then there is an element $a\in 1 + I$ such that $aM=0$.

Proof. Let $e_1,\ldots, e_n$ be a set of generators for $M$ (over $A$). Then we have that

$$e_1 = x_{11}e_1+ x_{12}e_2 + \cdots + x_{1n}e_n,$$

$$e_2 = x_{21}e_1+ x_{22}e_2 + \cdots + x_{2n}e_n,$$

$$\vdots \qquad \qquad \vdots \qquad \qquad \vdots \qquad \qquad \vdots$$

$$e_n = x_{n1}e_1+ x_{n2}e_2 + \cdots + x_{nn}e_n,$$

for $x_{ij} \in I$. Let $X$ be the matrix $(x_{ij})$, then the above set of equations is equivalent to

$$(X-\mathrm{id})\begin{pmatrix}e_1\\ e_2\\ \vdots\\ e_n\end{pmatrix} = 0,$$

where $\mathrm{id}$ is the identity matrix. Multiplying by the adjoint matrix of $X-\mathrm{id}$ we obtain that

$$\mathrm{det}(X-\mathrm{id})\begin{pmatrix}e_1\\ e_2\\ \vdots\\ e_n\end{pmatrix} = 0,$$

where $\mathrm{det}$ denotes determinant. Thus, $\mathrm{det}(X-\mathrm{id})e_i=0$ for all $i$, consequently $\mathrm{det}(X-\mathrm{id})M=0$. Expanding the determinant we see that $\mathrm{det}(X-\mathrm{id}) = 1+a$, with $a\in I$. $\qedsymbol$